# Is there a Definite Integral Representation for $n^n$?

The factorial $n!$ has a nice representation as definite integral: $$n!=\Gamma(n+1)=\int_0^\infty t^{n} e^{-t}\, \mathrm{d}t. \!$$ Is it possible to write down such an integral for $n^n$ as well?

I tried to come up with an integral that reproduces a $n$ factor, $n$-times, but without success. I don't see a way to stop the partial integration process like in the $n!$ case. So this might not work here and I currently can't think of another way. If it helps to restrict $n$, feel free to do so.

The only thing a found online so far is the Lambert's $W$ function, which is involved when solving $x^x=z$, but I'm not sure if this helps.

EDIT: Answers with integrals of the form $\displaystyle n^n=\int_0^\infty \cdots dt$ are preferred.

• Asymptotics like $\displaystyle n^n\approx \int_0^\infty (2\pi n)^{-1}(et)^{n} e^{-t}\, \mathrm{d}t$ don't count. Jul 26, 2012 at 8:47

No need for the Lambert W function , respecting your limits and using $\Gamma$ we get that

$$n^{n}=\frac{1}{\Gamma (n+1)}\int_{0}^{\infty }e^{-\frac{t^{ \frac{1}{n}}}{n}} dt$$

Update:

Lookup the Exponential Integral and its relationship with the Incomplete gamma function

$$E_{n}(x)=x^{n-1}\Gamma(1-n,x)\tag{1}$$

for $n=1-n$ and $x=\frac{1}{n}$ we get that :

$$E_{1-n}(\frac{1}{n})=\int_{1}^{\infty}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt = n^{n}\Gamma (n,\frac{1}{n})\tag{2}$$

Changing the integration limits to $[0,1]$ :

$$\int_{0}^{1}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt = n^{n}(\Gamma(n)-\Gamma (n,\frac{1}{n}))\tag{3}$$

Combining 2+3 we get the solution:

$$n^{n}=\frac{1}{\Gamma(n)}\int_{0}^{\infty}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt$$

The top result is due to the relationship :

$$n E_{1-n}(\frac{1}{n})= \int_{1}^{\infty} e^{-\frac{t^{\frac{1}{n}}}{n}} dt\tag{4}$$

• Thanks again, how did you get that? Jul 27, 2012 at 17:10
• @draks update how Jul 30, 2012 at 12:15
• Thanks again, great answer, vote up people$\color{green}{.}\color{goldenrod}{.}\color{red}{.}$ Jul 30, 2012 at 12:19

This may not be the line of thinking you are after, but it does give the integral you desire.

$$n^n=\int_0^n nx^{n-1}\ dx$$

• Thanks, no it's not. I thought about an infinite upper limit. But despite that it looks very interesting. +1 Jul 26, 2012 at 11:02

$$\frac{\Gamma (\alpha)}{s^{\alpha}}=\int_{0}^{\infty }t^{\alpha-1}e^{-st}dt\tag{1}$$

$$\frac{\Gamma (s)}{s^{s}}=\int_{0}^{\infty }t^{s-1}e^{-st}dt\tag{2}$$

$$s^{-s}=\frac{1}{\Gamma (s)}\int_{0}^{\infty }t^{s-1}e^{-st}dt\tag{3}$$

$$s^{s}=\frac{\Gamma (s)}{\int_{0}^{\infty }t^{s-1}e^{-st}dt}\tag{4}$$

• Hmm, not bad +1. invert it and we are even closer $\displaystyle s^{s}=\frac{\Gamma (s)}{\int_{0}^{\infty }t^{s-1}e^{-st}dt}$, at least on the lhs... Jul 26, 2012 at 11:05

The correct question (which someone has just asked me) may be as follows:

Find $f(t)$ (positive and independent of $n$) such that

$$n^n = \int_0^\infty t^n f(t) \,dt,\quad \forall\,n$$

That is, find $f(t)$ such that $\{n^n\}$ is the sequence of moments of f(t)dt. This is a particular instance of the Stieltjes moment problem.

There is criterion (due to Carleman) which guarantees that such a solution is unique, but I don't know an explicit form for $f(t)$.

• A rather strange edit of this post has been approved. However, it is possible that user65120 and the anonymous user, who suggested the edit, are the same person. Mar 5, 2013 at 12:30
• thanks for the background infos... Mar 5, 2013 at 14:28